Is the following system of equations solved using the singular value decomposition?

Question

Knowing the matrices $X$ and $Y$, I would like to solve the following system of equations for the vectors $u$ and $v$. \begin{gather} X^\top Y v - u u^\top X^\top Y v = 0\\ Y^\top X u - vv^\top Y^\top X u = 0, \end{gather} where $X, Y \in \mathbb{R}^{n \times p}$ and $u, v \in \mathbb{R}^p$ satisfy $\Vert u \Vert = \Vert v \Vert = 1$. I think if $Y = X$, then the solution is given by the top singular vectors. I'm not sure what to do in this case.

Answer 1

I think I figured it out myself, but please let me know if this is incorrect.

Let $M = X^\top Y$. Then $M^\top = Y^\top X$, and \begin{gather} Mv = \lambda u\\ M^\top u = \lambda v, \end{gather} where $\lambda = u^\top X^\top Y v = v^\top Y^\top X u$. In other words, $u$ is a left-singular vector of $M$, and $v$ is a right-singular vector of $M$.